3.349 \(\int \frac{\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

Optimal. Leaf size=120 \[ -\frac{\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{f (a+b)^3}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a f (a+b)^{7/2}}-\frac{\cot ^5(e+f x)}{5 f (a+b)}+\frac{(a+2 b) \cot ^3(e+f x)}{3 f (a+b)^2}-\frac{x}{a} \]

[Out]

-(x/a) + (b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*(a + b)^(7/2)*f) - ((a^2 + 3*a*b + 3*b^2)*Cot
[e + f*x])/((a + b)^3*f) + ((a + 2*b)*Cot[e + f*x]^3)/(3*(a + b)^2*f) - Cot[e + f*x]^5/(5*(a + b)*f)

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Rubi [A]  time = 0.343942, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {4141, 1975, 480, 583, 522, 203, 205} \[ -\frac{\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{f (a+b)^3}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a f (a+b)^{7/2}}-\frac{\cot ^5(e+f x)}{5 f (a+b)}+\frac{(a+2 b) \cot ^3(e+f x)}{3 f (a+b)^2}-\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

-(x/a) + (b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*(a + b)^(7/2)*f) - ((a^2 + 3*a*b + 3*b^2)*Cot
[e + f*x])/((a + b)^3*f) + ((a + 2*b)*Cot[e + f*x]^3)/(3*(a + b)^2*f) - Cot[e + f*x]^5/(5*(a + b)*f)

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{\cot ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{-5 (a+2 b)-5 b x^2}{x^4 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{5 (a+b) f}\\ &=\frac{(a+2 b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{-15 \left (a^2+3 a b+3 b^2\right )-15 b (a+2 b) x^2}{x^2 \left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{(a+b)^3 f}+\frac{(a+2 b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f}+\frac{\operatorname{Subst}\left (\int \frac{-15 (a+2 b) \left (a^2+2 a b+2 b^2\right )-15 b \left (a^2+3 a b+3 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^3 f}\\ &=-\frac{\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{(a+b)^3 f}+\frac{(a+2 b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a f}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a (a+b)^3 f}\\ &=-\frac{x}{a}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{a (a+b)^{7/2} f}-\frac{\left (a^2+3 a b+3 b^2\right ) \cot (e+f x)}{(a+b)^3 f}+\frac{(a+2 b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac{\cot ^5(e+f x)}{5 (a+b) f}\\ \end{align*}

Mathematica [C]  time = 2.9304, size = 671, normalized size = 5.59 \[ \frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (\csc (e) \csc ^5(e+f x) \left (540 a^2 b \sin (2 e+f x)-420 a^2 b \sin (2 e+3 f x)-240 a^2 b \sin (4 e+3 f x)+132 a^2 b \sin (4 e+5 f x)+225 a^2 b f x \cos (2 e+3 f x)-225 a^2 b f x \cos (4 e+3 f x)-45 a^2 b f x \cos (4 e+5 f x)+45 a^2 b f x \cos (6 e+5 f x)+780 a^2 b \sin (f x)+180 a^3 \sin (2 e+f x)-140 a^3 \sin (2 e+3 f x)-90 a^3 \sin (4 e+3 f x)+46 a^3 \sin (4 e+5 f x)+75 a^3 f x \cos (2 e+3 f x)-75 a^3 f x \cos (4 e+3 f x)-15 a^3 f x \cos (4 e+5 f x)+15 a^3 f x \cos (6 e+5 f x)+280 a^3 \sin (f x)+480 a b^2 \sin (2 e+f x)-400 a b^2 \sin (2 e+3 f x)-180 a b^2 \sin (4 e+3 f x)+116 a b^2 \sin (4 e+5 f x)+225 a b^2 f x \cos (2 e+3 f x)-225 a b^2 f x \cos (4 e+3 f x)-45 a b^2 f x \cos (4 e+5 f x)+45 a b^2 f x \cos (6 e+5 f x)+680 a b^2 \sin (f x)+150 f x (a+b)^3 \cos (2 e+f x)-150 f x (a+b)^3 \cos (f x)+75 b^3 f x \cos (2 e+3 f x)-75 b^3 f x \cos (4 e+3 f x)-15 b^3 f x \cos (4 e+5 f x)+15 b^3 f x \cos (6 e+5 f x)\right )-\frac{480 b^4 (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{\sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{960 a f (a+b)^3 \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*((-480*b^4*ArcTan[(Sec[f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b
)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(Sqrt
[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + Csc[e]*Csc[e + f*x]^5*(-150*(a + b)^3*f*x*Cos[f*x] + 150*(a + b)^3*f*
x*Cos[2*e + f*x] + 75*a^3*f*x*Cos[2*e + 3*f*x] + 225*a^2*b*f*x*Cos[2*e + 3*f*x] + 225*a*b^2*f*x*Cos[2*e + 3*f*
x] + 75*b^3*f*x*Cos[2*e + 3*f*x] - 75*a^3*f*x*Cos[4*e + 3*f*x] - 225*a^2*b*f*x*Cos[4*e + 3*f*x] - 225*a*b^2*f*
x*Cos[4*e + 3*f*x] - 75*b^3*f*x*Cos[4*e + 3*f*x] - 15*a^3*f*x*Cos[4*e + 5*f*x] - 45*a^2*b*f*x*Cos[4*e + 5*f*x]
- 45*a*b^2*f*x*Cos[4*e + 5*f*x] - 15*b^3*f*x*Cos[4*e + 5*f*x] + 15*a^3*f*x*Cos[6*e + 5*f*x] + 45*a^2*b*f*x*Co
s[6*e + 5*f*x] + 45*a*b^2*f*x*Cos[6*e + 5*f*x] + 15*b^3*f*x*Cos[6*e + 5*f*x] + 280*a^3*Sin[f*x] + 780*a^2*b*Si
n[f*x] + 680*a*b^2*Sin[f*x] + 180*a^3*Sin[2*e + f*x] + 540*a^2*b*Sin[2*e + f*x] + 480*a*b^2*Sin[2*e + f*x] - 1
40*a^3*Sin[2*e + 3*f*x] - 420*a^2*b*Sin[2*e + 3*f*x] - 400*a*b^2*Sin[2*e + 3*f*x] - 90*a^3*Sin[4*e + 3*f*x] -
240*a^2*b*Sin[4*e + 3*f*x] - 180*a*b^2*Sin[4*e + 3*f*x] + 46*a^3*Sin[4*e + 5*f*x] + 132*a^2*b*Sin[4*e + 5*f*x]
+ 116*a*b^2*Sin[4*e + 5*f*x])))/(960*a*(a + b)^3*f*(a + b*Sec[e + f*x]^2))

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Maple [A]  time = 0.112, size = 173, normalized size = 1.4 \begin{align*} -{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{fa}}-{\frac{1}{5\,f \left ( a+b \right ) \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}+{\frac{a}{3\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{2\,b}{3\,f \left ( a+b \right ) ^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{{a}^{2}}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}-3\,{\frac{ab}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}-3\,{\frac{{b}^{2}}{f \left ( a+b \right ) ^{3}\tan \left ( fx+e \right ) }}+{\frac{{b}^{4}}{f \left ( a+b \right ) ^{3}a}\arctan \left ({\tan \left ( fx+e \right ) b{\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( a+b \right ) b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x)

[Out]

-1/f/a*arctan(tan(f*x+e))-1/5/f/(a+b)/tan(f*x+e)^5+1/3/f/(a+b)^2/tan(f*x+e)^3*a+2/3/f/(a+b)^2/tan(f*x+e)^3*b-1
/f/(a+b)^3/tan(f*x+e)*a^2-3/f/(a+b)^3/tan(f*x+e)*a*b-3/f/(a+b)^3/tan(f*x+e)*b^2+1/f/(a+b)^3*b^4/a/((a+b)*b)^(1
/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.683559, size = 1972, normalized size = 16.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/60*(4*(23*a^3 + 66*a^2*b + 58*a*b^2)*cos(f*x + e)^5 - 20*(7*a^3 + 21*a^2*b + 20*a*b^2)*cos(f*x + e)^3 - 15
*(b^3*cos(f*x + e)^4 - 2*b^3*cos(f*x + e)^2 + b^3)*sqrt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4
- 2*(3*a*b + 4*b^2)*cos(f*x + e)^2 - 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sqrt(
-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*(a^3 + 3*
a^2*b + 3*a*b^2)*cos(f*x + e) + 60*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*
a*b^2 + b^3)*f*x*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^4 + 3*a^3*b + 3*a^2*
b^2 + a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^
2*b^2 + a*b^3)*f)*sin(f*x + e)), -1/30*(2*(23*a^3 + 66*a^2*b + 58*a*b^2)*cos(f*x + e)^5 - 10*(7*a^3 + 21*a^2*b
+ 20*a*b^2)*cos(f*x + e)^3 + 15*(b^3*cos(f*x + e)^4 - 2*b^3*cos(f*x + e)^2 + b^3)*sqrt(b/(a + b))*arctan(1/2*
((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*(a^3 + 3*a^2*b
+ 3*a*b^2)*cos(f*x + e) + 30*((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2
+ b^3)*f*x*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f*x)*sin(f*x + e))/(((a^4 + 3*a^3*b + 3*a^2*b^2 +
a*b^3)*f*cos(f*x + e)^4 - 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^4 + 3*a^3*b + 3*a^2*b^2
+ a*b^3)*f)*sin(f*x + e))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**6/(a+b*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.47196, size = 300, normalized size = 2.5 \begin{align*} \frac{\frac{15 \,{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b + b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )} \sqrt{a b + b^{2}}} - \frac{15 \,{\left (f x + e\right )}}{a} - \frac{15 \, a^{2} \tan \left (f x + e\right )^{4} + 45 \, a b \tan \left (f x + e\right )^{4} + 45 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 15 \, a b \tan \left (f x + e\right )^{2} - 10 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

1/15*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^4/((a^4 + 3*a^3*b +
3*a^2*b^2 + a*b^3)*sqrt(a*b + b^2)) - 15*(f*x + e)/a - (15*a^2*tan(f*x + e)^4 + 45*a*b*tan(f*x + e)^4 + 45*b^2
*tan(f*x + e)^4 - 5*a^2*tan(f*x + e)^2 - 15*a*b*tan(f*x + e)^2 - 10*b^2*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2
)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*tan(f*x + e)^5))/f